Solving polynominals equations (relationship of roots)Quadratic equation - $alpha$ and $beta$ RootsTechnique to simplify algebraic calculations on roots of polynomialInterval of Polynomial Root FindingFind $alpha^3 + beta^3$ which are roots of a quadratic equation.sum and product of roots of polynomials: finding equations for rootsSolving two Cubic Equation on their Roots.Finding an equation with related rootsFind the roots of $acx^2-b(c+a)x+(c+a)^2=0$If $3x^2-6x+p=0$ has roots $alpha$ and $beta$, then find a quadratic with roots $(alpha+beta)/alpha$ and $(alpha+beta)/beta$Find the roots of $3x^3-4x-8$
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Solving polynominals equations (relationship of roots)
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Solving polynominals equations (relationship of roots)
Quadratic equation - $alpha$ and $beta$ RootsTechnique to simplify algebraic calculations on roots of polynomialInterval of Polynomial Root FindingFind $alpha^3 + beta^3$ which are roots of a quadratic equation.sum and product of roots of polynomials: finding equations for rootsSolving two Cubic Equation on their Roots.Finding an equation with related rootsFind the roots of $acx^2-b(c+a)x+(c+a)^2=0$If $3x^2-6x+p=0$ has roots $alpha$ and $beta$, then find a quadratic with roots $(alpha+beta)/alpha$ and $(alpha+beta)/beta$Find the roots of $3x^3-4x-8$
$begingroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$
So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac113$
polynomials roots
edited 40 mins ago
Lee David Chung Lin
4,51851342
asked 1 hour ago
Alex Alex
186
$endgroup$
add a comment |
$begingroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$
So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac113$
polynomials roots
edited 40 mins ago
Lee David Chung Lin
4,51851342
asked 1 hour ago
Alex Alex
186
$endgroup$
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
add a comment |
$begingroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$
So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac113$
polynomials roots
edited 40 mins ago
Lee David Chung Lin
4,51851342
asked 1 hour ago
Alex Alex
186
$endgroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$
So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac113$
polynomials roots
polynomials roots
edited 40 mins ago
Lee David Chung Lin
4,51851342
asked 1 hour ago
Alex Alex
186
edited 40 mins ago
Lee David Chung Lin
4,51851342
asked 1 hour ago
Alex Alex
186
edited 40 mins ago
Lee David Chung Lin
4,51851342
edited 40 mins ago
Lee David Chung Lin
4,51851342
edited 40 mins ago
Lee David Chung Lin
4,51851342
4,51851342
asked 1 hour ago
Alex Alex
186
asked 1 hour ago
Alex Alex
186
asked 1 hour ago
Alex Alex
186
186
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
add a comment |
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
1
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$
$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$
$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$
$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$
$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
answered 1 hour ago
user1952500user1952500
1,5351016
$endgroup$
add a comment |
$begingroup$
Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
$endgroup$
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$
answered 16 mins ago
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$
$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$
$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$
$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$
$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
answered 1 hour ago
user1952500user1952500
1,5351016
$endgroup$
add a comment |
$begingroup$
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$
$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$
$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$
$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$
$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
answered 1 hour ago
user1952500user1952500
1,5351016
$endgroup$
add a comment |
$begingroup$
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$
$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$
$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$
$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$
$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
answered 1 hour ago
user1952500user1952500
1,5351016
$endgroup$
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$
$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$
$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$
$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$
$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
answered 1 hour ago
user1952500user1952500
1,5351016
answered 1 hour ago
user1952500user1952500
1,5351016
answered 1 hour ago
user1952500user1952500
1,5351016
answered 1 hour ago
user1952500user1952500
1,5351016
1,5351016
add a comment |
add a comment |
$begingroup$
Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
$endgroup$
add a comment |
$begingroup$
Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
$endgroup$
add a comment |
$begingroup$
Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
$endgroup$
Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
79.7k42867
add a comment |
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$
answered 16 mins ago
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$
answered 16 mins ago
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$
answered 16 mins ago
farruhotafarruhota
22.5k2942
$endgroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$
answered 16 mins ago
farruhotafarruhota
22.5k2942
answered 16 mins ago
farruhotafarruhota
22.5k2942
answered 16 mins ago
farruhotafarruhota
22.5k2942
answered 16 mins ago
farruhotafarruhota
22.5k2942
22.5k2942
add a comment |
add a comment |
$begingroup$
That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
$endgroup$
add a comment |
$begingroup$
That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
$endgroup$
add a comment |
$begingroup$
That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
$endgroup$
That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
14.7k3827
add a comment |
add a comment |
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$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago